Cho \(A=1+3+3^2+3^3+3^4+......+3^{2012}\) và \(B=3^{2013}:2\) TÍNH B-A
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Lời giải:
Ta có:
\(A-\frac{1}{2}=\frac{3}{2}+(\frac{3}{2})^2+...+(\frac{3}{2})^{2012}\)
\(\frac{3}{2}(A-\frac{1}{2})=(\frac{3}{2})^2+(\frac{3}{2})^3+....+(\frac{3}{2})^{2013}\\ \Rightarrow \frac{3}{2}(A-\frac{1}{2})-(A-\frac{1}{2})=(\frac{3}{2})^{2013}-\frac{3}{2}\)
$\Rightarrow \frac{1}{2}(A-\frac{1}{2})=(\frac{3}{2})^{2013}-\frac{3}{2}$
$A-\frac{1}{2}=2(\frac{3}{2})^{2013}-3$
$A=2(\frac{3}{2})^{2013}-2,5$
$\Rightarrow A-B=2(\frac{3}{2})^{2013}-2,5-(\frac{3}{2})^{2013}:2$
$=\frac{3}{2}(\frac{3}{2})^{2013}-2,5=(\frac{3}{2})^{2014}-2,5$
A=đã cho.
=>3A=3+3^2+3^3+3^4+...+3^2012+3^2013.
=>3A-A=3^2013-1.
=>2A=3^2013-1.
=>A=\(\frac{3^{2013-1}}{2}\)
=>B-A=3^2013:2-(3^2013-1)/82.
=>B-A=1/2.
Vậy B-A=1/2.
Ta có:
\(A=1+3+3^2+...+3^{2012}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{2013}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3+...+3^{2013}\right)-\left(1+3+3^2+...+3^{2012}\right)\)
\(\Rightarrow2A=3^{2013}-1\)
\(\Rightarrow A=\left(3^{2013}-1\right):2\)
Do \(B=3^{2013}:2\)
\(\Rightarrow B-A=3^{2013}:2-\left(3^{2013}-1\right):2\)
\(\Rightarrow B-A=\left(3^{2013}-3^{2013}+1\right):2\)
\(\Rightarrow B-A=1:2=\frac{1}{2}\)
Vậy \(B-A=\frac{1}{2}\)
\(3A=3+3^2+3^3+3^4+3^5+...+3^{2013}\)
\(A=\frac{3A-A}{2}=\frac{3^{2013}-1}{2}\)
\(B-A=\frac{3^{2013}}{2}-\frac{3^{2013}-1}{2}=\frac{1}{2}\)
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